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of having only #wo sources, we have nan. In this case, we would write the
expression for 4A? as the sum of a whole lot of amplitudes, complex numbers,
--- Trang 563 ---
squared, and we would get the square of each one, all added together, plus cross
terms bebween every pair, and if the cireumnstances are such that the latter average
out, then there will be no effects ofinterference. It may be that the various sources
are located in such random positions that, although the phase diference between
4s and 4s is also defnite, it is very different from that bebtween 4: and 4a, etc.
So we would get a whole lot of cosines, many plus, many minus, all averaging out.
So it is that in many circumstances we do not see the efects of interference,
but see only a collective, total intensity equal to the sum of all the intensities.
32-5 Scattering of light
The above leads us to an efect which occurs in air as a consequence of the
Irregular positions of the atoms. When we were discussing the index of refraction,
we saw that an incoming beam of light will make the atoms radiate again. The
electric fñeld of the incoming beam drives the electrons up and down, and they
radiate because of their acceleration. Phis scattered radiation combines to give
a beam in the same direction as the incoming beam, but of somewhat diferent
phase, and this is the origin of the index of refraction.
But what can we say about the amount of re-radiated light in some other
direction? Ordinarily, If the atoms are very beautifully located in a nice pattern,
1t is easy to show that we get nothing in other directions, because we are adding
a lot of vectors with their phases always changing, and the result comes to zero.
But ïf the objects are randomlụ located, then the total intensity in any direction
1s the sươn of the intensities that are scattered by each atom, as we have just
discussed. Eurthermore, the atoms in a gas are in actual motion, so that although
the relative phase of two atoms is a definite amount now, later the phase would be
quite diferent, and therefore eøch cosine term will average out. Therefore, to fnd
out how much light is scattered in a given direction by a gas, we merely study the
efects of one øtom and multiply the intensity it radiates by the number of atoms.
Barlier, we remarked that the phenomenon of scattering of light of this nature
1s the origin of the blue of the sky. 'Phe sunlight goes through the air, and when
we look to one side of the sun—say at 90° to the beam——we see blue light; what
we now have to calculate is hou rmụch light we see and 0h it 1s blue.
Tf the incident beam has the electric ñeld* E = oec”“f at the point where
the atom is located, we know that an electron in the atom will vibrate up and
- * When a Caret appears on a vector iÈ signifies that the componen‡s of the vector are complex:
#2 = (E„y, lụ, E„).
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lncident beam + Atom
(unpolarized) „ XS
Scattered ¬
Fig. 32-2. A beam of radiation falls on an atom and causes the
charges (electrons) in the atom to move. The moving electrons in turn
radiate In varlous directions.
down in response to this (Fig. 32-2). Erom Eq. (23.8), the response will be
Ê=—> “—-: (32.15)
m(uổ — 3 + iu)
W© could include the damping and the possibility that the atom acts like several
oscillators of diferent frequency and sum over the various frequencies, but for
simplicity let us just take one oscillator and neglect the damping. Then the
response to the external electric fñield, which we have already used in the calculation
of the index of refraction, is simply
&=—S—. (32.16)
m(duỗ — œ3)
We could now easily calculate the intensity of light that is emitted in various
đirections, using formula (32.2) and the acceleration corresponding to the above Z.
Rather than do this, however, we shall simply calculate the #ofal amownt oŸ
light scattered in ai/ directions, just to save time. 'Phe total amount of light
energy per second, scattered in all directions by the single atom, is oÝ course
given by Eaq. (32.6). 5o, putting together the various pieces and regrouping them,
W© getE
P = [(q2ø*/12meoe))qễ Eỗ Jmà(Ÿ — œ8)” ]
= (šeocE8)(8a/3)(q¿/16n2cm¿e")[j`/(ø2 — œ8)”]
= (šcocE8)(Smr3/3)|°/(Ÿ — w)'] (32.17)
for the total scattered power, radiated ín all directions.
--- Trang 565 ---
We have written the result in the above form because it is then easy %O
remember: First, the total energy that is scattered is proportional to the square
of the incident fñeld. What does that mean? Obviously, the square of the ineident
field is proportional to the energy which is coming in per second. In fact, the
energy incident per square meter per second is cọc times the average (H2) of
the square of the electric field, and If lo is the maximum value of #, then
(F2?) = šEÿ. In other words, the total energy scabtered is proportional to the
energy per square meter that comes in; the brighter the sunlight that is shining
in the sky, the brighter the sky is going to look.
Next, what ƒfraction oŸ the incoming light is scattered? Let us imagine a
“target” with a certain area, let us say ø, in the beam (not a real, material target,
because this would difract light, and so on; we mean an imaginary area drawn
in space). The total amount of energy that would pass through this surface ø
in a given circumstance is proportional both to the incoming intensity and to ơ,
and the total power would be
P= (š‹ạcE))ø. (32.18)
Now we invent an idea: we say that the atom scatters a total amount of
intensity which is the amount which would fall on a certain geometrical area,
and we give the answer by giving that area. That answer, then, is independent
of the incident intensity; it gives the ratio of the energy scattered to the energy
Ineident per square meter. In other words, the ratio
total energy scattered per second :
—————— san ørea.
energy incident per square meter per second
The signifcance of this area is that, if all the energy that impinged on that area
were to be spewed in all directions, then that is the amount of energy that would
be scattered by the atom.
This area is called a cross seclion for scaftering; the idea OoŸ cross section 1s
used constantly, whenever some phenomenon occurs in proportion to the intensity
of a beam. In such cases one always describes the amount of the phenomenon by
saying what the efective area would have to be to pick up that mụuch of the beam.
lt does not mean in any way that this oscillator actually has such an area. If
there were nothing present but a free electron shaking up and down there would